prove that the line segment joining the mid point of two sides of a triangle is parallel to the third side and half of it.
Answers
Hi
Let there be triangle ABC,
and midpoint P and Q
ΔABC~ΔAPQ
AB:AP=2:1
∠A is common
AC:AQ=2:1
Hence it is SAS similarity
We know that two triangles have same angles
∠B=∠P
∠C=∠Q
So, line BC and PQ are parallel to each other
Answer:
Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined .
To Prove : DE || BC and DE = 1 / 2 BC.
Const. : Produce the line segment DE to F , such that DE = EF. Join FC .
Proof : In △s AED and CEF, we have
AE = CE [∵E is the mid point of AC]
∠AED = ∠CEF[vert. opp.∠s]
and DE = EF [by construction]
∴ △AED ≅ △CEF [by SAS congruence axiom]
⇒ AD = CF ---(i)[c.p.c.t.]
and ∠ADE and ∠CEF ---(ii) [c.p.c.t.]
Now, D is the mid point of AB.
⇒ AD = DB ---(iii)
From (i) and (iii), CF = DB ---(iv)
Also, from (ii)
⇒ AD = || FC [if a pair of alt. int. ∠s are equal then lines are parallel]
⇒ DB || BC ---(v)
From (iv) and (v), we find that DBCF is a quadrilateral such that one pair of opposite sides are equal and parallel.
∴ DBCF is a ||gm
⇒ DF || BC and DF = BC [∵opp side of ||gm are equal and parallel]
Also, DE = EF [by construction]
Hence, DE || BC and DE = 1 / 2 BC
diagram is in the picture
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